定積分 $\int_2^3 \frac{x^2 + \sqrt{x^3} - \sqrt{x}}{x} dx$ を計算する。解析学定積分積分積分計算2025/7/71. 問題の内容定積分 ∫23x2+x3−xxdx\int_2^3 \frac{x^2 + \sqrt{x^3} - \sqrt{x}}{x} dx∫23xx2+x3−xdx を計算する。2. 解き方の手順まず、被積分関数を簡略化する。x2+x3−xx=x2x+x3x−xx=x+x3/2x−x1/2x=x+x1/2−x−1/2\frac{x^2 + \sqrt{x^3} - \sqrt{x}}{x} = \frac{x^2}{x} + \frac{\sqrt{x^3}}{x} - \frac{\sqrt{x}}{x} = x + \frac{x^{3/2}}{x} - \frac{x^{1/2}}{x} = x + x^{1/2} - x^{-1/2}xx2+x3−x=xx2+xx3−xx=x+xx3/2−xx1/2=x+x1/2−x−1/2したがって、積分は次のようになる。∫23(x+x1/2−x−1/2)dx\int_2^3 (x + x^{1/2} - x^{-1/2}) dx∫23(x+x1/2−x−1/2)dx各項を積分する。∫xdx=12x2\int x dx = \frac{1}{2}x^2∫xdx=21x2∫x1/2dx=23x3/2\int x^{1/2} dx = \frac{2}{3}x^{3/2}∫x1/2dx=32x3/2∫x−1/2dx=2x1/2\int x^{-1/2} dx = 2x^{1/2}∫x−1/2dx=2x1/2したがって、∫(x+x1/2−x−1/2)dx=12x2+23x3/2−2x1/2\int (x + x^{1/2} - x^{-1/2}) dx = \frac{1}{2}x^2 + \frac{2}{3}x^{3/2} - 2x^{1/2}∫(x+x1/2−x−1/2)dx=21x2+32x3/2−2x1/2定積分を計算する。∫23(x+x1/2−x−1/2)dx=[12x2+23x3/2−2x1/2]23\int_2^3 (x + x^{1/2} - x^{-1/2}) dx = \left[ \frac{1}{2}x^2 + \frac{2}{3}x^{3/2} - 2x^{1/2} \right]_2^3∫23(x+x1/2−x−1/2)dx=[21x2+32x3/2−2x1/2]23=(12(32)+23(33/2)−2(31/2))−(12(22)+23(23/2)−2(21/2))= \left(\frac{1}{2}(3^2) + \frac{2}{3}(3^{3/2}) - 2(3^{1/2})\right) - \left(\frac{1}{2}(2^2) + \frac{2}{3}(2^{3/2}) - 2(2^{1/2})\right)=(21(32)+32(33/2)−2(31/2))−(21(22)+32(23/2)−2(21/2))=(92+23(33)−23)−(2+23(22)−22)= \left(\frac{9}{2} + \frac{2}{3}(3\sqrt{3}) - 2\sqrt{3}\right) - \left(2 + \frac{2}{3}(2\sqrt{2}) - 2\sqrt{2}\right)=(29+32(33)−23)−(2+32(22)−22)=92+23−23−2−432+22= \frac{9}{2} + 2\sqrt{3} - 2\sqrt{3} - 2 - \frac{4}{3}\sqrt{2} + 2\sqrt{2}=29+23−23−2−342+22=92−2+22−432= \frac{9}{2} - 2 + 2\sqrt{2} - \frac{4}{3}\sqrt{2}=29−2+22−342=52+232= \frac{5}{2} + \frac{2}{3}\sqrt{2}=25+3223. 最終的な答え52+223\frac{5}{2} + \frac{2\sqrt{2}}{3}25+322